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== RockAndStones says ==
:::Gerhardvalentin, indeed assuming that you have a small insider's advantage, AND assuming almost symmetric host the conditional probability is close to 2/3. Einverstanden?[[User:RocksAndStones|RocksAndStones]] ([[User talk:RocksAndStones|talk]]) 18:45, 6 August 2011 (UTC)
::::Nein, RocksAndStones, not agreed, for that's the dodgy matter of dispute. Maths teachers like sundry "assumptions" as waste conditions and, according to their input forever will get then the adequate results. So that point is moot. For the "standard problem" all of this is not relevant to get the "only correct answer". You are right, assuming to have insider's advantage will result in ''probability to win by switching'' for the most extreme case of at least 1/2 (factor 2/3) and at the same time(!) likewise to 1 (factor 1/3). So always varying around 2/3.
::::What can conditional probability and superfluous assumption contribute to getting the only correct solution, the only correct answer, the only correct decision? Nothing but to confirm once more that nothing can be better than to switch here and now, in this one special game the question is about. That's no news at all.
::::Superfluous additional assumptions and their handling in conditional probability theory belong to the realm of probability theory. Solving the paradox of the MHP is easy, as anyone can clearly find out that two doors have double chance. And if you like you can add ''"have exactly double chance, because probability to win by switching will be at least 1/2 (factor 2/3) to even 1 (factor 1/3) in this one special game the question is about".'' That's all what conditional probability theory can contribute. So no forum here to teach and to learn conditional probability theory and to teach and to learn application of Bayes' theorem. That belongs to the maths' forum. Regards, [[User:Gerhardvalentin|Gerhardvalentin]] ([[User talk:Gerhardvalentin|talk]]) 11:14, 7 August 2011 (UTC)
::Gerhardvalentin, Perhaps, V R speaking about the same thing, so a "scheinbar" disagreement results from wording, instead of writing formulas. What I meant to say was: if Connie is a bit clairvoyant and can guess the worst door with chance say 0.35, her win probability is close to 2/3 no matter what Monte does. If we understand standard assumption as the uniform distribution, then always-switching gives 2/3 (no matter what Monte as door-opener does), although you really need an additional argument why Connie cannot do better (no matter what Monte does). That an argument is needed, can be seen with 4 doors, when Monte can help win with 100% probability by signaling through sequence of 2 doors to reveal. Now, as far as I know, Olle Haggstrom and Richard were first to focus on this Holy Grail argument. Combinatorially, the Holy Grail result means that if Connie's decision policy is not allowed to exploit the actual ___location of prize (which formally means she "does not know" where it was hidden) then she cannot win in all three cases out of three under any circumstances. I agree, that the site is not a forum for teaching the conditional probability. However, from the viewpoint of Markov decision processes (and other standpoints), the conditional probability is a natural quantification of the advantage of switching. Although the existence of cond. prob. is unnecessary assumption and its computation adds nothing to the right decision, it is a valuable thing. In particular, if the distribution of the prize is not uniform, and the first quess was not done in optimal way, you may decide for sticking on base of conditional probability. I do not object the cond. prob. approach, it only needs to be given a proper place as a tool to quantify the advantage.[[User:RocksAndStones|RocksAndStones]] ([[User talk:RocksAndStones|talk]]) 12:14, 7 August 2011 (UTC)
::::Rocks&Stones, R U interested in my view? Again, I M not fully einverstanden. Let us stick to the standard version, where the host not just offers to switch if you should have chosen the prize, etc... Let us stick to the standard version, where unproven "assumptions" of clairvoyance and opaque doors never will influence the decision asked for.
::::Yes, Richard's papers confirm that indeed there never will be any better decision than to switch here and now in this very special game the question is about, and moreover that there never will be any better decision than to switch in every one of these games, if this special game should ever be on stage in real life. I would like to show the chances for switching of 2/3 in "odds form" just at the beginning, also. Just to help convincing the reader. And later that the chances forever will remain within the very strict range of (at least!) 1/2 to even 1, and without given better knowledge forever exactly 2/3. But I M strictly against the totally unproven brazen assertion that the MHP is incomplete without the conditional calculus showing a variable representing the totally unknown "host's special behaviour" in this special game. Such variant should be treated where it belongs, in some maths forum, for it is not needed within the MHP article, and it does not belong to the MHP article. It is an interesting maths aspect, but is without any relevance to the question asked for and to the correct decision to switch. And the fulminant nonsensical history can be shown in the "history" section. Regards, [[User:Gerhardvalentin|Gerhardvalentin]] ([[User talk:Gerhardvalentin|talk]]) 13:57, 7 August 2011 (UTC)
::Gerhardvalentin, sure I am interested in your viewpoint and appreciate it. Let us review your statements word-by-word, to avoid ambiguity. ''Richard's papers confirm that indeed there never will be any better decision than to switch here'' I agree, vorausgesetzt uniform distribution, this was shown by Olle Haeggstroem (Lehrbuch Streifzuege der W-Theorie) and Richard by somewhat different arguments. ''I would like to show the chances for switching of 2/3 in "odds form" just at the beginning, also.'' If you mean unconditional odds - I agree this is easy and can be done at the very beginning, but if you mean conditional odds you need the assumption of symmetric host for the case when he has a freedom of choice. ''And later that the chances forever will remain within the very strict range of (at least!) 1/2 to even 1, and without given better knowledge forever exactly 2/3.'' I agree, under the assumption that host tosses a perhaps biased coin when he has the fredom of choice.''But I M strictly against the totally unproven brazen assertion that the MHP is incomplete without the conditional calculus showing a variable representing the totally unknown "host's special behaviour" in this special game. '' I agree with you at this point completely.''Such variant should be treated where it belongs, in some maths forum, for it is not needed within the MHP article, and it does not belong to the MHP article. It is an interesting maths aspect, but is without any relevance to the question asked for and to the correct decision to switch. And the fulminant nonsensical history can be shown in the "history" section. '' Well, the subject belongs to the MHP site, and the "history section" is indeed to where it belongs. Now, I have a straightforward question to you: have you ever taken 15 minutes to understand the dominance argument, explained since March at least half a dozen of times on these discussion pages?[[User:RocksAndStones|RocksAndStones]] ([[User talk:RocksAndStones|talk]]) 18:01, 7 August 2011 (UTC)
:::Thanks for your questions, RocksAndStones, I try to reply as follows
::#Uniform distribution of the three objects at the beginning is only affordable if the guest does not choose her door randomly. If she has no information on the distribution of the three objects and chooses her door uniformly at random, then a uniform distribution is not even necessary.
::#Richard's papers confirm that nothing can beat ''switching'' in each and every single game, ''no matter (!)'' whether "uniform distribution" or "biased host", and even more of that: The more biased, the better! In the end it never matters whether the host's method of showing one goat can tell you sth to revise the odds on the door first selected, or not! (By the way see also the [http://arxiv.org/PS_cache/arxiv/pdf/1105/1105.5809v1.pdf "Total Solution"] in ''"The Three Doors" by Sasha Gnedin'' May 31, 2011, e.g.: You can answer the question without needing probability at all). We should not stick on dated sources that do not even address the solution of that veridical paradox, but just use the "MHP" to present their competence and expertise in conditional probability theory. And textbooks written to teach conditional probability theory, using the MHP as a welcome example. No room here for such sources, showing correct calculi and explaining probability theory, but not the famous paradox.
::#The prior and the conditional probability can easily be shown in odds form, using Bayes' rule:
::::Prior odds:
::::Prob(C=1) : Prob(C=2) : Prob(C=3) = 1 : 1 : 1 (as long as no other information is available)
::::Likelihood of ___location of car C=1,2,3, given data Q=3:
::::Prob(Q=3|C=1) : Prob(Q=3|C=2) : Prob(Q=3|C=3) = 1/2 : 1: 0
::::Posterior odds:
::::Prob(C=1|Q=3) : Prob(C=2|Q=3) : Prob(C=3|Q=3) = 1/2 : 1 : 0
::::The posterior probabilities are therefore 1/3, 2/3, 0 since they must add to 1, and must be in the just mentioned ratios.
::::This computation is easy to generatlise to the situation where the host may cause asymmetry. As long as the initial distribution of the ___location of the car remains uniform, the posterior odds are
::::Prob(C=1|Q=3) : Prob(C=2|Q=3) : Prob(C=3|Q=3) = Prob(Q=3|C=1) : 1 : 0
::::As Prob(Q=3|C=1) can conceivably be anything between 0 and 1, the posterior odds on the ___location of the car being door 1 to door 2 is anything between 0 : 1 to 1 : 1. All we can say is "switching will never hurt".
:::As you can see, the assumption of a symmetric host – for the case that he has all the freedom of choice in the world – is not necessary at all, as switching will give you the car on average in 2 out of 3, and will ''never hurt.''
:::The host could be "assumed" to be (extremely) biased to open only the door with the lowest number e.g., whenever he can, i.e. if a goat is behind his preferred door. He can open his preferred door in two of three cases:
:::*if in 1/3 he has got two goats to show and switching will LOOSE the car
:::*if in 1/3 he has got the car and one goat, and the goat being behind his preferred door, and switching will WIN the car.
::::Whenever he opens his preferred door in 2 out of 3 cases, you know that the chance to win by switching is 1/2 (and never ever less).
:::*but if in the last 1/3 he has got one goat and the car, but the car being behind his preferred door, he has to open his avoided door with the higher number, and in this 1 out of 3 -case switching is very likely to win for sure.
:::Last item: No, didn't have the time yet to look to the dominance argument. Regards, [[User:Gerhardvalentin|Gerhardvalentin]] ([[User talk:Gerhardvalentin|talk]]) 22:07, 7 August 2011 (UTC)
:::Thank you, had a look there, and that's exactly the principle and affirmation to where any vain, though prominent, shiftless failed attempt imho irrevocably belongs to. [[User:Gerhardvalentin|Gerhardvalentin]] ([[User talk:Gerhardvalentin|talk]]) 16:43, 8 August 2011 (UTC)
== Conditional probability and the MHP ==
The MHP is a good example in teaching and learning conditional probability theory. But what is weight of conditional probability theory for solving the world famous paradox called "Monty Hall paradox"?
What "variant" of the MHP? It's about a "fair variant". Special presuppositions: at best "none".
The distribution of the three objects behind the three doors is of no relevance as long as it is ''unknown to the candidate'' who must assume 1/3:1/3:1/3.
Special preferences of the candidate in choosing one door are not relevant, nor the special behavior of the host in opening a door, as long as he observes the rules that say he has to open one door showing a goat and he has to offer his still closed door as an alternative. If he should dispose of two goats it's completely up to him which door to open. Because even if he should be extremely biased, to switch never can be of disadvantage to the candidate.
Whether the host should act extremely biased, or biased to a smaller extent, or not biased at all is without importance for this famous paradox. Morgan et al. however succeeded in luring almost everyone interested in this paradox on the wrong track. Conditional probability is OK as long as it is not misused to lure in the wrong direction in fixing the view on insignificance. The current scientific literature has helped here to provide clarity.
*What is "known"? The first known fact is the overall chance of winning by switching of 2/3. The chances of the "door first selected" to the "alternative still locked door" = "1/3 : 2/3". Agreed? And we can visualize an extremely biased host who, in opening his favored door, signals that switching is of no disadvantage for the candidate ("1/2 : 1/2" in two out of three) but who - in the last third of cases, in exceptionally opening his unfavored door, sigals that switching is very likely to win the car ("0/3 : 3/3" or "0 : 1").
*Once more: We know the "hypothetical" assumed worst case of "1/2 : 1/2" and the assumed best case of "0 : 1". Just assumptions, but a valid demarcation of the intermediate range. Not a wide range, but a fixed, closely limited range that categorically excludes that staying could ever be of advantage. And, besides assumptions, the only thing we really "know" is the exact overall probability of "1/3 : 2/3". So we can see that it is ''absolutely pointless'' to try to define hypothetical variants within this fixed, closely limited range, that never can advise that staying could be of advantage. Trying to define hypothetical variants by using conditional probability theory is absolutely meaningless. We know from the outset that this fixed, closely limited range forever discards staying, because, in the MHP, staying never can be the better decision.
*Conditional probability theory can never leave this fixed, closely limited range that – at the outset – categorically excludes that staying could ever be of advantage, and so never can be an option within the MHP.
It should be shown to the reader that in the MHP it is fruitless to use Bayes in search of variants within this narrow fixed range that, just from the outset, forever imperatively discards staying. [[User:Gerhardvalentin|Gerhardvalentin]] ([[User talk:Gerhardvalentin|talk]]) 22:07, 13 September 2011 (UTC)
=== Game Theory ===
[http://arxiv.org/PS_cache/arxiv/pdf/1105/1105.5809v1.pdf Gnedin has shown] in a solution using [[Game Theory]] that all strategies that consider "staying" as a possible option, and not imperatively decide "unexceptional switch" have to be discarded just from the outset. His solution shows that conditional probability, given the door numebers, or any assumptions regarding some special behavior of the host is completely irrelevant for the MHP. See [http://arxiv.org/PS_cache/arxiv/pdf/1105/1105.5809v1.pdf Gnedin: The Doors]. So I just said on [http://en.wikipedia.org/wiki/User_talk:Nijdam/Discussion#Game_theory Nijdam's talk page about the host's strategy] in his section [http://en.wikipedia.org/wiki/User_talk:Nijdam/Discussion#Game_theory "Game Theory"]:
::And yes Nijdam, the host indeed is able to determine his own strategy just as he should like. That's completely up to him.
::Moreover, it is completely irrelevant whether the candidate knows about such strategy, or not.
::Please note that any host's strategy is completely irrelevant for the decision asked for.
::All sources agree that staying never can be of advantage for the candidate and never can be the "better" decision.
::So any thought that considers "staying" is dominated and can be discarded just from the outset, as game theory can show and already did show.
Game theory indeed should be a topic in the article MHP, and I will be trying to write a draft including Game Theory. [[User:Gerhardvalentin|Gerhardvalentin]] ([[User talk:Gerhardvalentin|talk]]) 09:13, 14 September 2011 (UTC)
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